Simplify the following expression: $y = \dfrac{5x^2- 3x- 8}{5x - 8}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(-8)} &=& -40 \\ {a} + {b} &=& &=& {-3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-40$ and add them together. Remember, since $-40$ is negative, one of the factors must be negative. The factors that add up to ${-3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-8}$ and ${b}$ is ${5}$ $ \begin{eqnarray} {ab} &=& ({-8})({5}) &=& -40 \\ {a} + {b} &=& {-8} + {5} &=& -3 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({5}x^2 {-8}x) + ({5}x {-8}) $ Factor out the common factors: $ x(5x - 8) + 1(5x - 8)$ Now factor out $(5x - 8)$ $ (5x - 8)(x + 1)$ The original expression can therefore be written: $ \dfrac{(5x - 8)(x + 1)}{5x - 8}$ We are dividing by $5x - 8$ , so $5x - 8 \neq 0$ Therefore, $x \neq \frac{8}{5}$ This leaves us with $x + 1; x \neq \frac{8}{5}$.